A wide channel carries unit discharge q = 22.0 cfs/ft at a depth of 2.40 ft. Find the Froude number and critical depth.

Start with the thinking

• Fr > 1 is supercritical (fast, shallow).
• Critical depth for a wide channel is (q²/g)^(1/3).

Step-by-step solution

1. Velocity

   $$V = q/y = 22.0/2.40 = 9.17 ft/s$$

2. Froude

   $$Fr = V/\sqrt(gy)$$

3. Substituting

   $$Fr = 9.17/\sqrt(32.2 \times 2.40) = 1.043$$

4. Regime — supercritical

5. Critical depth

   $$y_c = (q^{2}/g)^(1/3)$$

6. Substituting

   $$y_c = ((22.0)^{2}/32.2)^(1/3) = 2.468 ft$$

Why the other options are there

• Fr = 0.119 (square root omitted)
• y_c = 15.03 ft (cube root omitted)

Reference: FE Reference Handbook — Hydraulics → Specific Energy Diagram

A wide channel carries a unit discharge of 15.0 cfs/ft at a depth of 4.75 ft. Compute the specific energy, the critical depth and the minimum specific energy.

Start with the thinking

• Specific energy is measured from the channel bottom: E = y + V²/2g.
• Minimum specific energy occurs at critical depth and equals 1.5 y_c for a rectangular section.

Step-by-step solution

1. Velocity

   $$V = q/y = 15.0/4.75 = 3.16 ft/s$$

2. Specific energy

   $$E = y + V^{2}/(2g)$$

3. Substituting

   $$E = 4.75 + (3.16)^{2}/64.4 = 4.905 ft$$

4. Critical depth

   $$y_c = (q^{2}/g)^(1/3)$$

5. Substituting

   $$y_c = ((15.0)^{2}/32.2)^(1/3) = 1.912 ft$$

6. Minimum energy

   $$E_min = 1.5 y_c$$

7. Substituting

   $$E_min = 1.5 \times 1.912 = 2.868 ft$$

8. Regime — the flow is subcritical because y > y_c

Why the other options are there

• 14.72 ft (2g omitted)
• 1.91 ft reported as E_min (1.5 factor dropped)

Reference: FE Reference Handbook — Hydraulics → Specific Energy Diagram

A rectangular channel carries a unit discharge of 3.5 m³/s per metre at a supercritical depth of 0.25 m. Compute the upstream Froude number, the sequent depth, and the energy dissipated in the jump.

Start with the thinking

• A jump only forms when the approach flow is supercritical (Fr > 1).
• The sequent-depth equation comes from momentum, while the loss comes from the energy equation.

Step-by-step solution

1. Velocity

   $$v_{1} = q/y_{1} = 3.5/0.25 = 14.000 m/s$$

2. Formula

   $$Fr_{1} = v_{1}/\sqrt(g y_{1})$$

3. Substituting

   $$Fr_{1} = 14.000/\sqrt(9.81 \times 0.25) = 8.940$$

4. Formula

   $$y_{2} = (y_{1}/2)[\sqrt(1 + 8Fr_{1}^{2}) - 1]$$

5. Substituting

   $$y_{2} = (0.25/2)[\sqrt(1 + 8 \times 8.940^{2}) - 1] = 3.038 m$$

6. Energies

   $$E_{1} = 0.25 + 9.990 = 10.240 m; E_{2} = 3.038 + 0.068 = 3.106 m$$

7. Loss — ΔE = 10.240 − 3.106 = 7.134 m

Why the other options are there

• y₂ = 2.23 m (depth simply scaled by Froude number)
• ΔE = 2.79 m (depth change reported as energy loss)

Reference: FE Reference Handbook — Hydraulics → Specific Energy Diagram

A wide channel carries unit discharge q = 14.5 cfs/ft at a depth of 4.40 ft. Find the Froude number and critical depth.

Start with the thinking

• Fr > 1 is supercritical (fast, shallow).
• Critical depth for a wide channel is (q²/g)^(1/3).

Step-by-step solution

1. Velocity

   $$V = q/y = 14.5/4.40 = 3.30 ft/s$$

2. Froude

   $$Fr = V/\sqrt(gy)$$

3. Substituting

   $$Fr = 3.30/\sqrt(32.2 \times 4.40) = 0.277$$

4. Regime — subcritical

5. Critical depth

   $$y_c = (q^{2}/g)^(1/3)$$

6. Substituting

   $$y_c = ((14.5)^{2}/32.2)^(1/3) = 1.869 ft$$

Why the other options are there

• Fr = 0.023 (square root omitted)
• y_c = 6.53 ft (cube root omitted)

Reference: FE Reference Handbook — Hydraulics → Specific Energy Diagram

A wide channel carries a unit discharge of 28.5 cfs/ft at a depth of 2.75 ft. Compute the specific energy, the critical depth and the minimum specific energy.

Start with the thinking

• Specific energy is measured from the channel bottom: E = y + V²/2g.
• Minimum specific energy occurs at critical depth and equals 1.5 y_c for a rectangular section.

Step-by-step solution

1. Velocity

   $$V = q/y = 28.5/2.75 = 10.36 ft/s$$

2. Specific energy

   $$E = y + V^{2}/(2g)$$

3. Substituting

   $$E = 2.75 + (10.36)^{2}/64.4 = 4.418 ft$$

4. Critical depth

   $$y_c = (q^{2}/g)^(1/3)$$

5. Substituting

   $$y_c = ((28.5)^{2}/32.2)^(1/3) = 2.933 ft$$

6. Minimum energy

   $$E_min = 1.5 y_c$$

7. Substituting

   $$E_min = 1.5 \times 2.933 = 4.399 ft$$

8. Regime — the flow is supercritical because y < y_c

Why the other options are there

• 110.2 ft (2g omitted)
• 2.93 ft reported as E_min (1.5 factor dropped)

Reference: FE Reference Handbook — Hydraulics → Specific Energy Diagram

A rectangular channel carries a unit discharge of 4.5 m³/s per metre at a supercritical depth of 0.45 m. Compute the upstream Froude number, the sequent depth, and the energy dissipated in the jump.

Start with the thinking

• A jump only forms when the approach flow is supercritical (Fr > 1).
• The sequent-depth equation comes from momentum, while the loss comes from the energy equation.

Step-by-step solution

1. Velocity

   $$v_{1} = q/y_{1} = 4.5/0.45 = 10.000 m/s$$

2. Formula

   $$Fr_{1} = v_{1}/\sqrt(g y_{1})$$

3. Substituting

   $$Fr_{1} = 10.000/\sqrt(9.81 \times 0.45) = 4.759$$

4. Formula

   $$y_{2} = (y_{1}/2)[\sqrt(1 + 8Fr_{1}^{2}) - 1]$$

5. Substituting

   $$y_{2} = (0.45/2)[\sqrt(1 + 8 \times 4.759^{2}) - 1] = 2.812 m$$

6. Energies

   $$E_{1} = 0.45 + 5.097 = 5.547 m; E_{2} = 2.812 + 0.131 = 2.943 m$$

7. Loss — ΔE = 5.547 − 2.943 = 2.604 m

Why the other options are there

• y₂ = 2.14 m (depth simply scaled by Froude number)
• ΔE = 2.36 m (depth change reported as energy loss)

Reference: FE Reference Handbook — Hydraulics → Specific Energy Diagram

A wide channel carries unit discharge q = 13.5 cfs/ft at a depth of 2.60 ft. Find the Froude number and critical depth.

Start with the thinking

• Fr > 1 is supercritical (fast, shallow).
• Critical depth for a wide channel is (q²/g)^(1/3).

Step-by-step solution

1. Velocity

   $$V = q/y = 13.5/2.60 = 5.19 ft/s$$

2. Froude

   $$Fr = V/\sqrt(gy)$$

3. Substituting

   $$Fr = 5.19/\sqrt(32.2 \times 2.60) = 0.567$$

4. Regime — subcritical

5. Critical depth

   $$y_c = (q^{2}/g)^(1/3)$$

6. Substituting

   $$y_c = ((13.5)^{2}/32.2)^(1/3) = 1.782 ft$$

Why the other options are there

• Fr = 0.062 (square root omitted)
• y_c = 5.66 ft (cube root omitted)

Reference: FE Reference Handbook — Hydraulics → Specific Energy Diagram

A wide channel carries a unit discharge of 22.5 cfs/ft at a depth of 2.00 ft. Compute the specific energy, the critical depth and the minimum specific energy.

Start with the thinking

• Specific energy is measured from the channel bottom: E = y + V²/2g.
• Minimum specific energy occurs at critical depth and equals 1.5 y_c for a rectangular section.

Step-by-step solution

1. Velocity

   $$V = q/y = 22.5/2.00 = 11.25 ft/s$$

2. Specific energy

   $$E = y + V^{2}/(2g)$$

3. Substituting

   $$E = 2.00 + (11.25)^{2}/64.4 = 3.965 ft$$

4. Critical depth

   $$y_c = (q^{2}/g)^(1/3)$$

5. Substituting

   $$y_c = ((22.5)^{2}/32.2)^(1/3) = 2.505 ft$$

6. Minimum energy

   $$E_min = 1.5 y_c$$

7. Substituting

   $$E_min = 1.5 \times 2.505 = 3.758 ft$$

8. Regime — the flow is supercritical because y < y_c

Why the other options are there

• 128.6 ft (2g omitted)
• 2.51 ft reported as E_min (1.5 factor dropped)

Reference: FE Reference Handbook — Hydraulics → Specific Energy Diagram

A rectangular channel carries a unit discharge of 8.0 m³/s per metre at a supercritical depth of 0.35 m. Compute the upstream Froude number, the sequent depth, and the energy dissipated in the jump.

Start with the thinking

• A jump only forms when the approach flow is supercritical (Fr > 1).
• The sequent-depth equation comes from momentum, while the loss comes from the energy equation.

Step-by-step solution

1. Velocity

   $$v_{1} = q/y_{1} = 8.0/0.35 = 22.857 m/s$$

2. Formula

   $$Fr_{1} = v_{1}/\sqrt(g y_{1})$$

3. Substituting

   $$Fr_{1} = 22.857/\sqrt(9.81 \times 0.35) = 12.335$$

4. Formula

   $$y_{2} = (y_{1}/2)[\sqrt(1 + 8Fr_{1}^{2}) - 1]$$

5. Substituting

   $$y_{2} = (0.35/2)[\sqrt(1 + 8 \times 12.335^{2}) - 1] = 5.933 m$$

6. Energies

   $$E_{1} = 0.35 + 26.628 = 26.978 m; E_{2} = 5.933 + 0.093 = 6.026 m$$

7. Loss — ΔE = 26.978 − 6.026 = 20.953 m

Why the other options are there

• y₂ = 4.32 m (depth simply scaled by Froude number)
• ΔE = 5.58 m (depth change reported as energy loss)

Reference: FE Reference Handbook — Hydraulics → Specific Energy Diagram

A wide channel carries unit discharge q = 22.5 cfs/ft at a depth of 3.60 ft. Find the Froude number and critical depth.

Start with the thinking

• Fr > 1 is supercritical (fast, shallow).
• Critical depth for a wide channel is (q²/g)^(1/3).

Step-by-step solution

1. Velocity

   $$V = q/y = 22.5/3.60 = 6.25 ft/s$$

2. Froude

   $$Fr = V/\sqrt(gy)$$

3. Substituting

   $$Fr = 6.25/\sqrt(32.2 \times 3.60) = 0.580$$

4. Regime — subcritical

5. Critical depth

   $$y_c = (q^{2}/g)^(1/3)$$

6. Substituting

   $$y_c = ((22.5)^{2}/32.2)^(1/3) = 2.505 ft$$

Why the other options are there

• Fr = 0.054 (square root omitted)
• y_c = 15.72 ft (cube root omitted)

Reference: FE Reference Handbook — Hydraulics → Specific Energy Diagram